1、题目

There are N gas stations along a circular route, where the amount of gas at station i is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.

Return the starting gas station’s index if you can travel around the circuit once, otherwise return -1.
环形公路加油站问题。一条环形公路上有N个加油站，加油站i的油量为gas[i]，从加油站i到i+1的耗油为cost[i]。汽车的油箱容量不限。求从哪个站点出发可以跑完全程，不存在则返回-1。

Note:
The solution is guaranteed to be unique.

2、分析

gas[i]-cost[i]等于从i到i+1还剩的油量。
int current 记录当前剩油量
int total 记录总剩油量
从0站点开始遍历：
对每站gas[i]-cost[i]求和。current+=gas[i]-cost[i]，total+=gas[i]-cost[i]
如果累计剩油量current<0，则0-i的所有站点都不可能作为出发点。(因为跑不到下一站就没油了）。把i+1设置为新出发点，current重置。

遍历结束后：如果total<0，说明总体是跑不到的，返回-1；否则返回找到的起始点。

3、代码

int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
    int total = 0;//总剩油量
    int current = 0;//当前剩油量
    int start = 0;
    for (int i = 0; i < gas.size(); i++) {
        current += gas[i] - cost[i];
        total += gas[i] - cost[i];
        if (current < 0)//i和之前的所有站点不能成为起始点
        {
            start = i + 1;
            current = 0;
        }
    }
    return total < 0 ? -1 : start;
}